Calculus I - Finding Absolute Extrema

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Section 4.4 : Finding Absolute Extrema

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13. Determine the absolute extrema of \(R\left( x \right) = \ln \left( {{x^2} + 4x + 14} \right)\) on \(\left[ { - 4,2} \right]\).

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First, notice that we are working with a logarithm whose argument is a polynomial (which is continuous everywhere) that is always positive in the interval. Because of this we can see that the function will be continuous on the given interval. Recall that this is important because we now know that absolute extrema will in fact exist by the Extreme Value Theorem!

Now that we know that absolute extrema will in fact exist on the given interval we’ll need to find the critical points of the function.

Given that the purpose of this section is to find absolute extrema we’ll not be putting much work/explanation into the critical point steps. If you need practice finding critical points please go back and work some problems from that section.

Here are the critical points for this function.

\[R'\left( x \right) = \frac{{2x + 4}}{{{x^2} + 4x + 14}}\hspace{0.5in} \Rightarrow \hspace{0.5in}\,\,x = - 2\] Show Step 2

Now, recall that we actually are only interested in the critical points that are in the given interval and so, in this case, the critical point that we need is,

\[x = - 2\] Show Step 3

The next step is to evaluate the function at the critical points from the second step and at the end points of the given interval. Here are those function evaluations.

\[R\left( { - 4} \right) = 2.6391\hspace{0.5in}R\left( { - 2} \right) = 2.3026\hspace{0.5in}R\left( 2 \right) = 3.2581\] Show Step 4

The final step is to identify the absolute extrema. So, the answers for this problem are then,

\[\require{bbox} \bbox[2pt,border:1px solid black]{\begin{align*}{\mbox{Absolute Maximum : }} & 3{\mbox{.2581 at }}x = 2\\ {\mbox{Absolute Minimum : }} & 2.3026{\mbox{ at }}x = - 2\end{align*}}\]